The feeders in battling machine are represented in proportions and fractions.
- The equation that represents the problem is:
- The feeder can hold <em>30 baseballs</em>, when full
The given parameters are:
<em />
<em> ------ 1/6 full</em>
<em />
<em> --- baseballs added</em>
<em />
<em> ---- 2/3 full</em>
<em />
So, the equation that represents the problem is:
So, we have:
The number of baseballs it can hold is calculated as follows:
Multiply through by 6
Collect like terms
Divide through by 3
Hence, the feeder can hold 30 baseballs, when full
Read more about proportions and fractions at:
brainly.com/question/20337104
Let x represent the side length of the square end, and let d represent the dimension that is the sum of length and girth. Then the volume V is given by
V = x²(d -4x)
Volume will be maximized when the derivative of V is zero.
dV/dx = 0 = -12x² +2dx
0 = -2x(6x -d)
This has solutions
x = 0, x = d/6
a) The largest possible volume is
(d/6)²(d -4d/6) = 2(d/6)³
= 2(108 in/6)³ = 11,664 in³
b) The dimensions of the package with largest volume are
d/6 = 18 inches square by
d -4d/6 = d/3 = 36 inches long
Sense there's only a y, you'd assume y is 1 or 1y.
you'd make the 1 a negative 1 to get rid of the subtraction sign.
you'd add together the 2 ys making it 2y. then add -1 + 3, which would be 2.
so, the answer would be 2y + 2.
Answer:
Step-by-step explanation:
Given
comparing it with standard equation 
so 4a=1
so Focus of parabola is (0,0)
directrix
y=-a
here 
